The reaction $\mathrm { NO } ( g ) + \mathrm { O } _ { 3 } ( g ) \longrightarrow \mathrm { NO } _ { 2 } ( g ) + \mathrm { O } _ { 2 } ( g )$ was studied by performing two experiments. In the first experiment (results shown in following table), the rate of disappearance of NO was followed in a large excess of $\mathrm { O } _ { 3 . }$ (The $\left[ \mathrm { O } _ { 3 } \right]$ remains effectively constant at $1.0 \times 10 ^ { 14 }$ molecules/cm $^ { 3 } . )$

$$\begin{array} { c c } { \text { Time } } & { [ \mathrm { NO } ] } \\ { \text { (ms) } } & { \text { (molecules/cm } ^ { 3 } ) } \\ \hline { 0 } & { 6.0 \times 10 ^ { 8 } } \\ { 100 \pm 1 } & { 5.0 \times 10 ^ { 8 } } \\ { 500 \pm 1 } & { 2.4 \times 10 ^ { 8 } } \\ { 700 \pm 1 } & { 1.7 \times 10 ^ { 8 } } \\ { 1000 \pm 1 } & { 9.9 \times 10 ^ { 7 } } \end{array}$$

In the second experiment, $[ \mathrm { NO } ]$ was held constant at $2.0 \times 10 ^ { 14 }$ molecules/cm $^ { 3 } .$ The data for the disappearance of $\mathrm { O } _ { 3 }$ were as follows:

$$\begin{array} { c c } { \text { Time } } & { \left[ \mathrm { O } _ { 3 } \right] } \\ { ( \mathrm { ms } ) } & { \text { (molecules/cm } ^ { 3 } ) } \\ \hline { 0 } & { 1.0 \times 10 ^ { 10 } } \\ { 50 \pm 1 } & { 8.4 \times 10 ^ { 9 } } \\ { 100 \pm 1 } & { 7.0 \times 10 ^ { 9 } } \\ { 200 \pm 1 } & { 4.9 \times 10 ^ { 9 } } \\ { 300 \pm 1 } & { 3.4 \times 10 ^ { 9 } } \end{array}$$

What is the value of the rate constant obtained from each set of experiments? $\text {Rate} = k ^ { \prime } [ \mathrm { NO } ] ^ { x } \quad \text {Rate} = k ^ { \prime \prime } \left[ \mathrm { O } _ { 3 } \right] ^ { y }.$

The reaction $\mathrm { NO } ( g ) + \mathrm { O } _ { 3 } ( g ) \longrightarrow \mathrm { NO } _ { 2 } ( g ) + \mathrm { O } _ { 2 } ( g )$ was studied by performing two experiments. In the first experiment (results shown in following table), the rate of disappearance of NO was followed in a large excess of $\mathrm { O } _ { 3 . }$ (The $\left[ \mathrm { O } _ { 3 } \right]$ remains effectively constant at $1.0 \times 10 ^ { 14 }$ molecules/cm $^ { 3 } . )$

$$\begin{array} { c c } { \text { Time } } & { [ \mathrm { NO } ] } \\ { \text { (ms) } } & { \text { (molecules/cm } ^ { 3 } ) } \\ \hline { 0 } & { 6.0 \times 10 ^ { 8 } } \\ { 100 \pm 1 } & { 5.0 \times 10 ^ { 8 } } \\ { 500 \pm 1 } & { 2.4 \times 10 ^ { 8 } } \\ { 700 \pm 1 } & { 1.7 \times 10 ^ { 8 } } \\ { 1000 \pm 1 } & { 9.9 \times 10 ^ { 7 } } \end{array}$$

In the second experiment, $[ \mathrm { NO } ]$ was held constant at $2.0 \times 10 ^ { 14 }$ molecules/cm $^ { 3 } .$ The data for the disappearance of $\mathrm { O } _ { 3 }$ were as follows:

$$\begin{array} { c c } { \text { Time } } & { \left[ \mathrm { O } _ { 3 } \right] } \\ { ( \mathrm { ms } ) } & { \text { (molecules/cm } ^ { 3 } ) } \\ \hline { 0 } & { 1.0 \times 10 ^ { 10 } } \\ { 50 \pm 1 } & { 8.4 \times 10 ^ { 9 } } \\ { 100 \pm 1 } & { 7.0 \times 10 ^ { 9 } } \\ { 200 \pm 1 } & { 4.9 \times 10 ^ { 9 } } \\ { 300 \pm 1 } & { 3.4 \times 10 ^ { 9 } } \end{array}$$

What is the overall rate law?

The reaction $\mathrm { NO } ( g ) + \mathrm { O } _ { 3 } ( g ) \longrightarrow \mathrm { NO } _ { 2 } ( g ) + \mathrm { O } _ { 2 } ( g )$ was studied by performing two experiments. In the first experiment (results shown in following table), the rate of disappearance of NO was followed in a large excess of $\mathrm { O } _ { 3 . }$ (The $\left[ \mathrm { O } _ { 3 } \right]$ remains effectively constant at $1.0 \times 10 ^ { 14 }$ molecules/cm $^ { 3 } . )$

$$\begin{array} { c c } { \text { Time } } & { [ \mathrm { NO } ] } \\ { \text { (ms) } } & { \text { (molecules/cm } ^ { 3 } ) } \\ \hline { 0 } & { 6.0 \times 10 ^ { 8 } } \\ { 100 \pm 1 } & { 5.0 \times 10 ^ { 8 } } \\ { 500 \pm 1 } & { 2.4 \times 10 ^ { 8 } } \\ { 700 \pm 1 } & { 1.7 \times 10 ^ { 8 } } \\ { 1000 \pm 1 } & { 9.9 \times 10 ^ { 7 } } \end{array}$$

In the second experiment, $[ \mathrm { NO } ]$ was held constant at $2.0 \times 10 ^ { 14 }$ molecules/cm $^ { 3 } .$ The data for the disappearance of $\mathrm { O } _ { 3 }$ were as follows:

$$\begin{array} { c c } { \text { Time } } & { \left[ \mathrm { O } _ { 3 } \right] } \\ { ( \mathrm { ms } ) } & { \text { (molecules/cm } ^ { 3 } ) } \\ \hline { 0 } & { 1.0 \times 10 ^ { 10 } } \\ { 50 \pm 1 } & { 8.4 \times 10 ^ { 9 } } \\ { 100 \pm 1 } & { 7.0 \times 10 ^ { 9 } } \\ { 200 \pm 1 } & { 4.9 \times 10 ^ { 9 } } \\ { 300 \pm 1 } & { 3.4 \times 10 ^ { 9 } } \end{array}$$

What is the value of the rate constant for the overall rate law? $\text {Rate} = k [ \mathrm { NO } ] ^ { x } [ \mathrm { O } _ { 3 } ] ^ { y }.$

(3.75)(2.4)

Experimental values for the temperature dependence of the rate constant for the gas-phase reaction $\mathrm { NO } ( g ) + \mathrm { O } _ { 3 } ( g ) \longrightarrow \mathrm { NO } _ { 2 } ( g ) + \mathrm { O } _ { 2 } ( \mathrm { g } )$ are as follows:

$$\begin{array} { l l } { T ( \mathrm { K } ) } & { k \left( \mathrm { L } \operatorname { mol } ^ { - 1 } \mathrm { s } ^ { - 1 } \right) }\\ \hline \\ { 195 } & { 1.08 \times 10 ^ { 9 } } \\ { 230 . } & { 2.95 \times 10 ^ { 9 } } \\ { 260 . } & { 5.42 \times 10 ^ { 9 } } \\ { 298 } & { 12.0 \times 10 ^ { 9 } } \\ {369} & { 35.5 \times 10 ^ { 9 } } \end{array}$$

Make the appropriate graph using these data, and determine the activation energy for this reaction.

Jill had $24. She spent$8. Then how much money did Jill have left? Use a subtraction pattern to solve the problem. write an equation and find the answer.

The function c(p) = 0.99p represents the cost in dollars of p pounds of peaches. If the cost per pound increases by 10%, how will the graph of the function change? A. Translation 0.1 unit up B. Translation 0.1 unit right C. Horizontal stretch by a factor of 1.1 D. Vertical stretch by a factor of 1.1.

An approximation to the function $\sin x$ is $\sin x \approx x - x ^ { 3 } / 6 .$ Plot the $\sin x$ function and 20 evenly spaced error bars representing the error in the approximation.

Write an equation for the x-axis.

Find all real solutions of the equation.

$$\frac { 1 } { x - 1 } - \frac { 2 } { x ^ { 2 } } = 0$$

The 60th term of an arithmetic sequence is 106.5, and the common difference is 1.5. What is the first term of the sequence?

Use the properties of inverse functions to find the exact value of the expression, if possible. $\arctan \left(\tan \frac{11 \pi}{6}\right)$

A reaction of the form

$$\mathrm{aA} \longrightarrow \text { Products }$$

gives a plot of $\ln [\mathrm{A}]$ versus time (in seconds), which is a straight line with a slope of $-7.35 \times 10^{-3}.$ Assuming $[\mathrm{A}]_{0}=0.0100\ M,$ calculate the time (in seconds) required for the reaction to reach 22.9% completion.

An equilibrium mixture at 2000 K, 1 atm consists of $\mathrm{Cs}, \mathrm{Cs}^{+}$ and $e^-.$ Based on 1 kmol of Cs present initially, determine the percent ionization of cesium. At 2000 K, the ionization-equilibrium constant for $\mathrm{Cs} \rightleftarrows \mathrm{Cs}^{+}+\mathrm{e}^{-}$ is K = 15.63.